\xiti
\begin{enhancedline}

\begin{xiaotis}

\xiaoti{$x$ 取什么数时，下列分式有意义？}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={10em, colsep=0pt}}
        \xxt{$\dfrac{x}{3x - 1}$；} & \xxt{$\dfrac{x^2}{x + 1}$；} & \xxt{$\dfrac{3}{0.5x - 1}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{在下列各分式中，当 $x$ 等于什么数时，分式的值是零？当 $x$ 等于什么数时，分式没有意义？}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{2x + 1}{2 - x}$；} & \xxt{$\dfrac{2x - 0.5}{3x + 1}$。}
    \end{tblr}

\end{xiaoxiaotis}


\xiaoti{写出下列各等式中未知的分子或分母：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{2}{xy} = \dfrac{?}{x^2y^2}$；} & \xxt{$\dfrac{3x}{x + y} = \dfrac{?}{5(x + y)^2}$；} \\
        \xxt{$\dfrac{a + b}{ma + mb} = \dfrac{1}{?}$；} & \xxt{$\dfrac{(x - 1)^2}{x^2 - 1} = \dfrac{?}{x + 1}$；} \\
        \xxt{$\dfrac{x^2 + xy + y^2}{x^3 - y^3} = \dfrac{?}{x - y}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{不改变分式的值，把下列各式的分子与分母中的各项系数都化为整数：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{0.01x - 0.5}{0.3x + 0.04}$；} & \xxt{$\dfrac{2a - \dfrac{3}{2}b}{\dfrac{2}{3}a + b}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{不改变分式的值，而使分母的第一项系数是正数，下面的做法对不对？如果不对，应当怎样改正？}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{-a + b}{-a - b} = \dfrac{a + b}{a - b}$；} & \xxt{$\dfrac{1}{-x + y} = -\dfrac{1}{x + y}$；} \\
    \end{tblr}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{-x + 1}{-x - 1} = \dfrac{x - 1}{x + 1}$。}
    \end{tblr}

\end{xiaoxiaotis}


\xiaoti{不改变分式本身的符号和分式的值，使下列各组里第二个分式的分母和第一个分式的分母相同：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{1 + x}{x - 1}$，$\dfrac{2x}{1 - x}$；} & \xxt{$\dfrac{6x + 1}{x^2 - x + 3}$，$\dfrac{4x - 5}{-x^2 + x - 3}$；} \\
        \xxt{$\dfrac{3x}{(x - 1)(x - 2)}$，$\dfrac{3 + x}{(x - 1)(2 - x)}$；} & \xxt{$\dfrac{a}{(a - b)(b - c)}$，$\dfrac{b}{(b - a)(c - b)}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{约分：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={10em, colsep=0pt}}
        \xxt{$\dfrac{-xy^2}{xy}$；} & \xxt{$\dfrac{4a^2b}{-2a}$；} & \xxt{$\dfrac{a - b}{2(a - b)}$；} \\
        \xxt{$\dfrac{-8x^2y^2}{-12x^4y}$；} & \xxt{$\dfrac{12a^3b^4}{15a^3b^2c}$；} & \xxt{$\dfrac{a^2 + 3ab}{a^2b + 3ab^2}$；}
    \end{tblr}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$\dfrac{x^2 - 2x - 3}{x^3 + 2x^2 - 15x}$；} & \xxt{$\dfrac{2x^3 - 2y^3}{4x^2 + 4xy + 4y^2}$；} \\
        \xxt{$\dfrac{b^3 + b}{b^3 - 2b^2 + b - 2}$；} & \xxt{$\dfrac{(a - b)(b - c)(c - a)}{(b - a)(a - c)(c - b)}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{先约分，再求值：}
\begin{xiaoxiaotis}

    \xxt{$\dfrac{x^3 + y^3}{x^3 - x^2y + xy^2}$，其中 $x = 5$，$y = 3.5$；}

    \xxt{$\dfrac{3a^2 - ab}{9a^2 - 6ab + b^2}$，其中 $a = \dfrac{3}{4}$，$b = -\dfrac{2}{3}$。}

\end{xiaoxiaotis}

\xiaoti{计算：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}, rows={rowsep+=.25em}}
        \xxt{$\dfrac{2}{x} \div \dfrac{4}{x}$；} & \xxt{$8a^2b^4 \cdot \dfrac{-3a}{4b^3}$；} \\
        \xxt{$\dfrac{4a^4b^2}{15n^3} \div \dfrac{-8a^2b^2}{35n^2}$；} & \xxt{$\dfrac{a^2 - 4b^2}{3ab^2} \cdot \dfrac{ab}{a - 2b}$；} \\
        \xxt{$\dfrac{x^2 - y^2}{x} \cdot \dfrac{-x^2}{(x + y)^3}$；} & \xxt{$\left(\dfrac{y^2}{-x^3}\right)^4$；} \\
        \xxt{$\left(\dfrac{b^2}{a^3}\right)^n$ （$n$ 为正整数）；} & \xxt{$\left(\dfrac{b^{n+1}}{a^n}\right)^2$ （$n$ 为正整数）；} \\
        \xxt{$\left(\dfrac{x^2y}{-z^2}\right)^3 \div \left(\dfrac{-x^3}{z}\right)^2$；} & \xxt{$\left(\dfrac{a^2b}{-c}\right)^3 \cdot \left(\dfrac{c^2}{-ab}\right)^2 \div \left(\dfrac{bc}{a}\right)^4$；} \\
        \xxt{$\dfrac{x^2 - y^2}{xy} \div (x - y)$；} & \xxt{$(xy - x^2) \div \dfrac{x^2 - 2xy + y^2}{xy} \cdot \dfrac{x - y}{x^2}$；} \\
        \xxt{$\dfrac{x^2 + 2x + 4}{x^2 + 4x + 4} \div \dfrac{x^3 - 8}{3x + 6} \div \dfrac{1}{x^2 - 4}$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{两个圆柱的底面半径分别是 $r_1$ 厘米和 $r_2$ 厘米，高都是 $h$ 厘米。求它们的体积的比。}

\end{xiaotis}

\end{enhancedline}
%\vspace*{1em}

